We can bound the number of maximal \(\alpha \) -gapped palindromes by similar proofs to \(28 \alpha n + 7 n\) .
This bound solves an open problem in [1]}, where Kolpakov and Kucherov
conjectured that the number of \(\alpha \) -gapped palindromes with \(\alpha \ge 2\)  in a string is linear.
We briefly explain the main differences and similarities needed to understand the relationship between
gapped repeats and palindromes.
Let \(\sigma \)  be a maximal \(\alpha \) -gapped repeat (or \(\alpha \) -gapped palindrome).
If the gapped repeat (palindrome) has a periodic prefix \({s}_{{\lambda }}\)  generated by some run, the right arm has a periodic prefix (suffix) \({s}_{{\rho }}\) 
generated by a run of the same period.
Since \(\sigma \)  is maximal, both runs have to obey constraints that are similar in both cases, considering whether \(\sigma \)  is a gapped repeat or a gapped palindrome.
So it is easy to change the proof of lemmaP in order to work with palindromes.
Like with aperiodic gapped repeats,
we can apply the point analysis to the aperiodic \(\alpha \) -gapped palindromes, too.
As main idea, we map a gapped palindrome \({u}_{{\lambda }},v,{u}_{{\rho }}\)  injectively to the pair of integers \((\mathsf {e}({u}_{{\lambda }}), {v})\) ,
exchanging the period with the size of the gap. See appendix for proofs.
